Question

which functions have removable discontinuities (holes)? choose three correct answers.
$f(x)=\frac{x^{2}+4x + 4}{x^{2}+2x - 8}$
$f(x)=\frac{x^{2}-9}{x^{2}+7x + 12}$
$f(x)=\frac{x + 7}{x^{2}+5x - 14}$
$f(x)=\frac{x - 1}{x^{2}-1}$

Explanation:

Step1: Factor numerator and denominator

• For \(f(x)=\frac{x^{2}+4x + 4}{x^{2}+2x - 8}\):
• Numerator: \(x^{2}+4x + 4=(x + 2)^{2}\)
• Denominator: \(x^{2}+2x - 8=(x + 4)(x - 2)\)
• No common factors.
• For \(f(x)=\frac{x^{2}-9}{x^{2}+7x + 12}\):
• Numerator: \(x^{2}-9=(x + 3)(x - 3)\)
• Denominator: \(x^{2}+7x + 12=(x + 3)(x + 4)\)
• Common factor \((x + 3)\).
• For \(f(x)=\frac{x + 7}{x^{2}+5x - 14}\):
• Numerator: \(x + 7\)
• Denominator: \(x^{2}+5x - 14=(x + 7)(x - 2)\)
• Common factor \((x + 7)\).
• For \(f(x)=\frac{x - 1}{x^{2}-1}\):
• Numerator: \(x - 1\)
• Denominator: \(x^{2}-1=(x + 1)(x - 1)\)
• Common factor \((x - 1)\).

Answer:

\(f(x)=\frac{x^{2}-9}{x^{2}+7x + 12}\), \(f(x)=\frac{x + 7}{x^{2}+5x - 14}\), \(f(x)=\frac{x - 1}{x^{2}-1}\)